Giải thích các bước giải:

\(\begin{array}{l}
6,\\
\sin 2x = 2\sin x.\cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
6{\sin ^2}x + 2{\sin ^2}2x = 5\\
 \Leftrightarrow 6{\sin ^2}x + 2.{\left( {2\sin x.\cos x} \right)^2} = 5\\
 \Leftrightarrow 6{\sin ^2}x + 2.4{\sin ^2}x.{\cos ^2}x - 5 = 0\\
 \Leftrightarrow 6{\sin ^2}x + 8{\sin ^2}x.{\cos ^2}x - 5 = 0\\
 \Leftrightarrow 6{\sin ^2}x + 8{\sin ^2}x.\left( {1 - {{\sin }^2}x} \right) - 5 = 0\\
 \Leftrightarrow 6{\sin ^2}x + 8{\sin ^2}x - 8{\sin ^4}x - 5 = 0\\
 \Leftrightarrow  - 8{\sin ^4}x + 14{\sin ^2}x - 5 = 0\\
 \Leftrightarrow 8{\sin ^4}x - 14{\sin ^2}x + 5 = 0\\
 \Leftrightarrow \left( {4{{\sin }^2}x - 5} \right)\left( {2{{\sin }^2}x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
4{\sin ^2}x - 5 = 0\\
2{\sin ^2}x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = \dfrac{5}{4}\\
{\sin ^2}x = \dfrac{1}{2}
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1\\
 \Rightarrow {\sin ^2}x = \dfrac{1}{2} \Leftrightarrow \sin x =  \pm \dfrac{{\sqrt 2 }}{2} \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
7,\\
2{\sin ^4}x + 2{\cos ^4}x = 2\sin 2x - 1\\
 \Leftrightarrow 2.\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = 2\sin 2x - 1\\
 \Leftrightarrow 2.\left[ {\left( {{{\sin }^4}x + 2{{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right) - 2{{\sin }^2}x.{{\cos }^2}x} \right] = 2\sin 2x - 1\\
 \Leftrightarrow 2.\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2.{{\sin }^2}x.{{\cos }^2}x} \right] = 2\sin 2x - 1\\
 \Leftrightarrow 2.\left[ {{1^2} - 2.{{\sin }^2}x.{{\cos }^2}x} \right] = 2\sin 2x - 1\\
 \Leftrightarrow 2 - 4{\sin ^2}x.{\cos ^2}x = 2\sin 2x - 1\\
 \Leftrightarrow 2 - {\left( {2\sin x.\cos x} \right)^2} = 2\sin 2x - 1\\
 \Leftrightarrow 2 - {\sin ^2}2x = 2\sin 2x - 1\\
 \Leftrightarrow {\sin ^2}2x + 2\sin 2x - 3 = 0\\
 \Leftrightarrow \left( {\sin 2x + 3} \right)\left( {\sin 2x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x + 3 = 0\\
\sin 2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x =  - 3\\
\sin 2x = 1
\end{array} \right.\\
 - 1 \le \sin 2x \le 1 \Rightarrow \sin 2x = 1\\
 \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \\
8,\\
9 - 13\cos x + \dfrac{4}{{1 + {{\tan }^2}x}} = 0\\
 \Leftrightarrow 9 - 13\cos x + \dfrac{4}{{1 + \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = 0\\
 \Leftrightarrow 9 - 13\cos x + \dfrac{4}{{\dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}}} = 0\\
 \Leftrightarrow 9 - 13\cos x + \dfrac{{4{{\cos }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x}} = 0\\
 \Leftrightarrow 9 - 13\cos x + \dfrac{{4{{\cos }^2}x}}{1} = 0\\
 \Leftrightarrow 4{\cos ^2}x - 13\cos x + 9 = 0\\
 \Leftrightarrow \left( {\cos x - 1} \right)\left( {4\cos x - 9} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
4\cos x - 9 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{9}{4}
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = 1 \Rightarrow x = 2k\pi \\
9,\\
5.\left( {1 + \cos x} \right) = 2 + {\sin ^4}x - {\cos ^4}x\\
 \Leftrightarrow 5 + 5\cos x = 2 + \left( {{{\sin }^2}x - {{\cos }^2}x} \right).\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
 \Leftrightarrow 5 + 5\cos x = 2 + \left( {{{\sin }^2}x - {{\cos }^2}x} \right).1\\
 \Leftrightarrow 5 + 5\cos x = 2 + {\sin ^2}x - {\cos ^2}x\\
 \Leftrightarrow 5 + 5\cos x = 2 + \left( {1 - {{\cos }^2}x} \right) - {\cos ^2}x\\
 \Leftrightarrow 5 + 5\cos x = 3 - 2{\cos ^2}x\\
 \Leftrightarrow 2{\cos ^2}x + 5\cos x + 2 = 0\\
 \Leftrightarrow \left( {2\cos x + 1} \right)\left( {\cos x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\cos x + 1 = 0\\
\cos x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - \dfrac{1}{2}\\
\cos x =  - 2
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x =  - \dfrac{1}{2} \Leftrightarrow x =  \pm \dfrac{{2\pi }}{3} + k2\pi \\
10,\\
2{\sin ^2}x - 3\sin x + 1 = 0\\
 \Leftrightarrow \left( {2\sin x - 1} \right)\left( {\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin x - 1 = 0\\
\sin x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{1}{2}\\
\sin x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
x \in \left[ {0;\dfrac{\pi }{2}} \right) \Rightarrow x = \dfrac{\pi }{6}\\
Vậy\,\,\,\,phương\,\,\,trình\,\,\,đã\,\,\,\,cho\,\,\,\,có\,\,\,\,1\,\,\,\,nghiệm\,\,\,trên\,\,\,\left[ {0;\dfrac{\pi }{2}} \right)\\
11,\,\,\,\,2\cos 2x + 2\cos x - \sqrt 2  = 0\\
 \Leftrightarrow 2.\left( {2{{\cos }^2}x - 1} \right) + 2\cos x - \sqrt 2  = 0\\
 \Leftrightarrow 4{\cos ^2}x - 2 + 2\cos x - \sqrt 2  = 0\\
 \Leftrightarrow {\left( {2\cos x} \right)^2} - {\sqrt 2 ^2} + 2\cos x - \sqrt 2  = 0\\
 \Leftrightarrow \left( {2\cos x - \sqrt 2 } \right)\left( {2\cos x + \sqrt 2 } \right) + \left( {2\cos x - \sqrt 2 } \right) = 0\\
 \Leftrightarrow \left( {2\cos x - \sqrt 2 } \right)\left( {2\cos x + \sqrt 2  + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\cos x - \sqrt 2  = 0\\
2\cos x + \sqrt 2  + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 2 }}{2}\\
\cos x =  - \dfrac{{\sqrt 2  + 1}}{2}
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = \dfrac{{\sqrt 2 }}{2} \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k2\pi \\
x =  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\\
x \in \left[ {0;3\pi } \right] \Rightarrow x \in \left\{ {\dfrac{\pi }{4};\dfrac{{7\pi }}{4};\dfrac{{9\pi }}{4}} \right\}\\
Tổng\,\,\,tất\,\,\,cả\,\,\,các\,\,\,\,nghiệm\,\,\,\,của\,\,\,phương\,\,\,\,trình\,\,\,\,đã\,\,\,cho\,\,\,\,là:\,\,T = \dfrac{\pi }{4} + \dfrac{{7\pi }}{4} + \dfrac{{9\pi }}{4} = \dfrac{{17\pi }}{4}\\
12,\,\,\,\,4{\sin ^2}2x - 2\left( {1 + \sqrt 2 } \right)\sin 2x + \sqrt 2  = 0\\
 \Leftrightarrow 4{\sin ^2}2x - 2\sin 2x - 2\sqrt 2 \sin 2x + \sqrt 2  = 0\\
 \Leftrightarrow \left( {4{{\sin }^2}2x - 2\sin 2x} \right) - \left( {2\sqrt 2 \sin 2x - \sqrt 2 } \right) = 0\\
 \Leftrightarrow 2\sin 2x\left( {2\sin 2x - 1} \right) - \sqrt 2 \left( {2\sin 2x - 1} \right) = 0\\
 \Leftrightarrow \left( {2\sin 2x - 1} \right)\left( {2\sin 2x - \sqrt 2 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin 2x - 1 = 0\\
2\sin 2x - \sqrt 2  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \dfrac{1}{2}\\
\sin 2x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{{5\pi }}{6} + k2\pi \\
2x = \dfrac{\pi }{4} + k2\pi \\
2x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{{5\pi }}{{12}} + k\pi \\
x = \dfrac{\pi }{8} + k\pi \\
x = \dfrac{{3\pi }}{8} + k\pi 
\end{array} \right.\\
x \in \left( {0;\pi } \right) \Rightarrow x \in \left\{ {\dfrac{\pi }{{12}};\dfrac{{5\pi }}{{12}};\dfrac{\pi }{8};\dfrac{{3\pi }}{8}} \right\}\\
 \Rightarrow Có\,\,\,\,4\,\,\,\,nghiệm\,\,\,\,của\,\,\,phương\,\,\,\,trình\,\,\,đã\,\,\,cho\,\,\,trên\,\,\,\,\left( {0;\pi } \right)\\
13,\,\,\,\,{\sin ^2}2x - \cos 2x + 1 = 0\\
 \Leftrightarrow \left( {1 - {{\cos }^2}2x} \right) - \cos 2x + 1 = 0\\
 \Leftrightarrow  - {\cos ^2}2x - \cos 2x + 2 = 0\\
 \Leftrightarrow {\cos ^2}2x + \cos 2x - 2 = 0\\
 \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {\cos 2x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x - 1 = 0\\
\cos 2x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 1\\
\cos 2x =  - 2
\end{array} \right.\\
 - 1 \le \cos 2x \le 1 \Rightarrow \cos 2x = 1\\
 \Leftrightarrow 2x = k2\pi  \Leftrightarrow x = k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
x \in \left[ { - \pi ;4\pi } \right] \Rightarrow x \in \left\{ { - \pi ;0;\pi ;2\pi ;3\pi ;4\pi } \right\}\\
Vậy\,\,\,phương\,\,\,trình\,\,\,đã\,\,\,cho\,\,\,có\,\,\,6\,\,\,nghiệm\,\,\,thoả\,\,\,mãn
\end{array}\)